Simplify and expand the following expression: $ \dfrac{3}{4x - 24}+ \dfrac{4}{2x - 18}- \dfrac{x}{x^2 - 15x + 54} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{3}{4x - 24} = \dfrac{3}{4(x - 6)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{4}{2x - 18} = \dfrac{4}{2(x - 9)}$ We can factor the quadratic in the third term: $ \dfrac{x}{x^2 - 15x + 54} = \dfrac{x}{(x - 6)(x - 9)}$ Now we have: $ \dfrac{3}{4(x - 6)}+ \dfrac{4}{2(x - 9)}- \dfrac{x}{(x - 6)(x - 9)} $ The least common multiple of the denominators is: $ 8(x - 6)(x - 9)$ In order to get the first term over $8(x - 6)(x - 9)$ , multiply by $\dfrac{2(x - 9)}{2(x - 9)}$ $ \dfrac{3}{4(x - 6)} \times \dfrac{2(x - 9)}{2(x - 9)} = \dfrac{6(x - 9)}{8(x - 6)(x - 9)} $ In order to get the second term over $8(x - 6)(x - 9)$ , multiply by $\dfrac{4(x - 6)}{4(x - 6)}$ $ \dfrac{4}{2(x - 9)} \times \dfrac{4(x - 6)}{4(x - 6)} = \dfrac{16(x - 6)}{8(x - 6)(x - 9)} $ In order to get the third term over $8(x - 6)(x - 9)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{x}{(x - 6)(x - 9)} \times \dfrac{8}{8} = \dfrac{8x}{8(x - 6)(x - 9)} $ Now we have: $ \dfrac{6(x - 9)}{8(x - 6)(x - 9)} + \dfrac{16(x - 6)}{8(x - 6)(x - 9)} - \dfrac{8x}{8(x - 6)(x - 9)} $ $ = \dfrac{ 6(x - 9) + 16(x - 6) - 8x} {8(x - 6)(x - 9)} $ Expand: $ = \dfrac{6x - 54 + 16x - 96 - 8x}{8x^2 - 120x + 432} $ $ = \dfrac{14x - 150}{8x^2 - 120x + 432}$ Simplify: $ = \dfrac{7x - 75}{4x^2 - 60x + 216}$